13 nonnormality

8/3/2019 13 Nonnormality

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presented by:

Non-Normality

Dudi Barmana, M.Si.

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Agenda

y Konsekuensi yang akan dihadapi

y Identifikasi/pendeteksian (pemeriksaaan pola

sisaan dan uji-uji formal)

y Beberapa alternatif solusi: Transformasi

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Today Quote

Orang sering melempar batu di jalan kita. Tergantung kita mau

membuat batu itu jadi tembok atau jembatan

---Chinese book of wisdom---

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Konsekuensi yangakan dihadapi


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y If the errors come from a distribution with

thicker or heavier tails than the normal, LS fit

may be sensitive to a small subset of the data.

y Heavy-tailed error distributions often generate

outliers that pull LS fit too much in their

direction.

y Prediction could be invalid.y IndividualT-Test and Model F-Test could be

missleading.

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Identifikasi/pendeteksian


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Residual Plot

yGraphical analysis is a very effective way to

investigate the adequacy of the fit of a

regressio

n mo

del

and to

check theunderlying assumption.

yNormal Probability Plot:

y

Normal probability plot: a simple way to checkthe normal assumption.


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y A straight plot is indicative of normality

y A shape is indicative of right skew errors

(eg gamma)

y A shape is indicative of a symmetric,short-tailed distribution

y A shape is indicative of a symmetric long-tailed distribution


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y Ranked residuals: e[1] < < e[n]

y Plot e[i] against Pi = (i-1/2)/ny Sometimes plot e[i] against *

-1[ (i-1/2)/n]

y Plot nearly a straight line for large sample n > 32

if e[i] normaly Small sample (n


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Kolmogorov-Smirnov Test

Let be the cdf for the distribution.x)P(XF(x) i e!

In the uniform(0,1) case: 1x0x,F(x) ee!

Compare this to the empirical distribution function:

x)sampletheinX(#n

1(x)F in e!

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If X1, X2, , Xn really come from the distribution with cdf F, the distance

F(x)-(x)FmaxDD nx

n!!

should be small.

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Computing the test statistic:

Suppose we simulate 7 uniform(0,1)s and get:

0.6 0.2 0.5 0.9 0.1 0.4 0.2

(obviously simplified)

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Put them in order:

0.1xfor0(x)F7 !


0.6 0.2 0.5 0.9 0.1 0.4 0.2

0.6 0.2 0.5 0.9 0.1 0.4 0.2

Now the empirical cdf is:

0.2x0.1for7

1(x)F7 e!

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0.6 0.2 0.5 0.9 0.1 0.4 0.2

0.2x0.1for7

1(x)F7 e!

0.4x0.2for7

3(x)F7 e!

0.5x0.4for7

4(x)F

7

e!

0.6x0.5for7

5(x)F7 e!

0.9x0.6for7

6(x)F7 e!

0.9xfor1(x)F7 u!

0.1xfor0(x)F7 !

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0.6 0.2 0.5 0.9 0.1 0.4 0.2

0.257142935

9D7 }!

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Let X(1), X(2), ,X(n) be the ordered sample.

_ a-nnn D,DmaxD !Then Dn can be estimated by

where

! ee

)F(X-n

imaxD (i)ni1n

!ee

n

1-i-)F(XmaxD (i)

ni1n

(assuming non-repeating values)

This is exact

for the

uniform

distribution!

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We reject that this sample came from the proposed distribution if the

empirical cdf is too far from the true cdf of the proposed

distribution

ie: We reject if Dn is too large.

ie: How large is large?

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In the 1930s, Kolmogorov and Smirnov showed that

g

!gp

!e1i

t2i--1in

1/2

n

22e(-1)21-tDnPlim

So, for large sample sizes, you could assume

g

!

}e1i

t2i--1in

1/2 22e(-1)21-tDnP

and find the value of t that makes the right hand side

for an level test.E1-

E

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For small samples, people have worked out and tabulated critical values,

but there is no nice closed form solution.

J. Pomeranz (1973)

J . Durbin (1968)

n

1.6276

n

1.5174

n

1.3581

n

1.2239

n

1.0730cv

0.010.020.050.100.20E

Good approximations for n>40:

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For our small sample of size 7,

From a table, the critical value for a 0.05 level test for n=7 is 0.483.

0.257142935

9D7 }!

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For our large sample of size 100,000,

20.00152392D100000 !

The approximate critical value for a 0.05 level test for n=100,000 is

90.00429468100000

1.3518}


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Bera and Jarque testing

y It can be proved that the coefficients of skewness and kurtosis

can be expressed respectively as:

and

y

The Bera Jarque test statistic is given by

y We estimate b1

and b2 using the residuals from the OLS

regression, .u

b

E u1

3

2 3 2!

[ ]/

W b

E u2

4

22

![ ]

W

2~24

3

6

2

2

2

2

1 G

-

!

bbTW


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Beberapa alternatif solusi:

Transformasi


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Transformation on y:TheBox-Cox Method

(Power transformation: y

)

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y Box and Cox (1964) show how the parameters of the

regression model and P can be estimated simultaneously

using the method of maximum likelihood.y Use

Where is the geometric

mean of the observations and fit the model

y is related to the Jocobian of the transformation

converting the response variable y into

]ln/1[ln1

1 !

!

n

i

iyny

IFP

! Xy1P

y)(P

y


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y Computation Procedure:

y Choose P to minimize SSRes()

y Use 10-20 values ofP to compute SSRes(). Then plotSSRes() v.s. P. Finally read the value ofP that minimizes

SSRes() from graph.

y A second iteration can be performed using a finer mesh of

values if desired.y Cannot select Pby directly comparing residual sum of

squares from the regressions of on x because of a

different scale.

yOnce P is selected, the analyst is free to fit the model usingy (P { 0)or ln y (P = 0).

)(Py


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y AnApproximate Confidence Interval for P

y The C.I. can be useful in selecting the final value for P.

y For example: if the 0.596 is the minimizing value ofSSRes(), but if 0.5 is in the C.I., then we would prefer

choose P = 0.5. If 1 is in the C.I., then no transformationmay be necessary.

y Maximize

y An approximate 100(1-E)% C.I. for P is


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y Let

y can be approximated by

or where R is the number ofresidual degrees of freedom.

y This is based on

y exp(x) = 1 + x + x2/2! +

y

)/exp( 21, nEG nz /1

2

2/E

n/1 2 1,EGnt /1 2 ,2/ RE

222

1 RG tz $!


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13 nonnormality

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